Theorem sbalex 2235

Description: Equivalence of two ways to express proper substitution of a setvar for another setvar disjoint from it in a formula. This proof of their equivalence does not use df-sb 2068.

That both sides of the biconditional express proper substitution is proved by sb5 2268 and sb6 2088. The implication "to the left" is equs4v 2003 and does not require ax-10 2137 nor ax-12 2171. It also holds without disjoint variable condition if we allow more axioms (see equs4 2416). Theorem 6.2 of [Quine] p. 40. Theorem equs5 2460 replaces the disjoint variable condition with a distinctor antecedent. Theorem equs45f 2459 replaces the disjoint variable condition on 𝑥, 𝑡 with the nonfreeness hypothesis of 𝑡 in 𝜑. (Contributed by NM, 14-Apr-2008.) Revised to use equsexv 2260 in place of equsex 2418 in order to remove dependency on ax-13 2372. (Revised by BJ, 20-Dec-2020.) Revise to remove dependency on df-sb 2068. (Revised by BJ, 21-Sep-2024.)

Assertion

Ref	Expression
sbalex	⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑡 → 𝜑))

Distinct variable group:   𝑥,𝑡

Allowed substitution hints:   𝜑(𝑥,𝑡)

Proof of Theorem sbalex

Step	Hyp	Ref	Expression
1		nfa1 2148	. . 3 ⊢ Ⅎ𝑥∀𝑥(𝑥 = 𝑡 → 𝜑)
2		ax12v2 2173	. . . 4 ⊢ (𝑥 = 𝑡 → (𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜑)))
3	2	imp 407	. . 3 ⊢ ((𝑥 = 𝑡 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑡 → 𝜑))
4	1, 3	exlimi 2210	. 2 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑡 → 𝜑))
5		equs4v 2003	. 2 ⊢ (∀𝑥(𝑥 = 𝑡 → 𝜑) → ∃𝑥(𝑥 = 𝑡 ∧ 𝜑))
6	4, 5	impbii 208	1 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑡 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396  ∀wal 1537  ∃wex 1782

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-10 2137  ax-12 2171

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-ex 1783  df-nf 1787

This theorem is referenced by:  equsexvOLD  2261  sb5  2268  dfsb7  2276  mopick  2627  alexeqg  3581  dfdif3  4049  pm13.196a  42032